Explain the process of converting an equation of the form $y^2=x^3+a\,x+b$ to the equation $Y^2\,Z=X^3+a\,X\,Z^2+b\,Z^3$
This is projective coordinates. The idea is to express $x$ and $y$ as ratios $\frac XZ$ and $\frac YZ$ (with $x$, $y$, $X$, $Y$, $Z$ all in the base field). This has the beneficial effect that during point multiplication, we can defer to the end the expensive step of modular inversion that occurs for each point addition and doubling when we use Cartesian coordinates $(x,y)$.
Starting from from $y^2=x^3+a\,x+b$, changing $x$ to $\frac XZ$ and $y$ to $\frac YZ$ yields $\frac{Y^2}{Z^2}=\frac{X^3}{Z^3}+a\frac XZ+b$. And then multiplying by $Z^3$ yields $Y^2\,Z=X^3+a\,X\,Z^2+b\,Z^3$.
To obtain explicit point addition and doubling formulas, we can transform the standard formulas by making the same change of variable. There are many ways to arrange the computations, see the Explicit-Formulas Database.
To convert from $(x,y)$ to $(X,Y,Z)$, set $X=x$, $Y=y$, and $Z=1$.
To convert from $(X,Y,Z)$ to $(x,y)$ when $Z\ne0$, compute the inverse $Z^{-1}$ of $Z$, then set $x={Z^{-1}}X$ and $y={Z^{-1}}y$.
Note: We can initially represent the neutral/point at infinity $\mathcal O$ as $(X,Y,Z)=(0,1,0)$, and these formulas yield correct results, including when $\mathcal O$ is obtained as an intermediary value. At the end of a sequence of operations, we can convert back $(X,Y,Z)$ to $\mathcal O$ when $Z=0$.
